Add the quantities ${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{sin}\mathit{\omega }\mathit{t}$, ${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{15}\mathbf{sin}\left(\omega t+30°\right)$and${\mathit{y}}_{\mathbf{3}}\mathbf{=}\mathbf{5}\mathbf{sin}\left(\omega t-45°\right)$ using the phasor method

The equation of first wave is$y_1=10\sin \omega t$

The equation of second wave is $y_2=15\sin \left(\omega t+30°\right)$

The equation of third wave is $y_3=5\sin \left(\omega t-45°\right)$

The amplitude of the resultant wave is equal to the vector sum of the amplitude of each wave.

The horizontal component of the resultant wave is given as:

$$\begin{gathered} y_h=10\cos 0°+15\cos 30°+5c\mathrm{os}\left(-45°\right) \\ {y}_h=10+13+3.54 \\ {y}_h=26.54 \end{gathered}$$

The vertical component of the resultant wave is given as:

$$\begin{gathered} y_v=10\sin 0°+15\sin 30°+5\sin \left(-45°\right) \\ {y}_v=3.96 \end{gathered}$$

The resultant amplitude of waves is given as:

$$\begin{gathered} y_r=\sqrt{y_h^2+y_v^2} \\ {y}_r=\sqrt{{\left(26.54\right)}^2+{\left(3.96\right)}^2} \\ {y}_r=26.83 \end{gathered}$$

The direction of the resultant wave is given as:

$\begin{array}{rcl}\tan \theta & =& \frac{y_v}{y_h}\\ \tan \theta & =& \frac{3.96}{26.54}\\ \theta & =& 8.5°\end{array}$

The sum of the wave is given as:

$y=y_r\sin \left(\omega t+\theta \right)$

Substitute all the values in equation.

$y=\left(26.83\right)\sin \left(\omega t+8.5°\right)$

Therefore, the sum of wave is $\left(26.83\right)\sin \left(\omega t+8.5°\right)$.